Difference between revisions of "1995 IMO Problems/Problem 2"
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By Cauchy-Schwarz, <cmath>\left(\sum_{cyc}\dfrac{1}{a^3 (b+c)}\right)\left(\sum_{cyc}a(b+c)\right)\geq \left( \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right) ^2=(ab+ac+bc)^2</cmath> Dividing by <math>2(ab+bc+ac)</math> gives <cmath>\dfrac{1}{a^3 (b+c)}+\dfrac{1}{b^3 (a+c)}+\dfrac{1}{c^3 (a+b)}\geq \dfrac{1}{2}(ab+bc+ac)\geq \dfrac{3}{2}</cmath> by AM-GM. | By Cauchy-Schwarz, <cmath>\left(\sum_{cyc}\dfrac{1}{a^3 (b+c)}\right)\left(\sum_{cyc}a(b+c)\right)\geq \left( \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right) ^2=(ab+ac+bc)^2</cmath> Dividing by <math>2(ab+bc+ac)</math> gives <cmath>\dfrac{1}{a^3 (b+c)}+\dfrac{1}{b^3 (a+c)}+\dfrac{1}{c^3 (a+b)}\geq \dfrac{1}{2}(ab+bc+ac)\geq \dfrac{3}{2}</cmath> by AM-GM. | ||
+ | |||
+ | === Solution 4 === | ||
+ | Proceed as in Solution 1, to arrive at the equivalent inequality | ||
+ | <cmath> \frac{x^2}{y+z} + \frac{y^2}{z+x} + \frac{z^2}{x+y} \ge \frac{3}{2} . </cmath> | ||
+ | But we know that <cmath>x + y + z \ge 3xyz \ge 3</cmath> by AM-GM. Furthermore, | ||
+ | <cmath> (x + y + y + z + x + z) (\frac{x^2}{y+z} + \frac{y^2}{z+x} + \frac{z^2}{x+y}) \ge (x + y + z)^2 </cmath> | ||
+ | by Cauchy-Schwarz, and so dividing by <math>2(x + y + z)</math> gives | ||
+ | <cmath> | ||
+ | as desired. | ||
{{alternate solutions}} | {{alternate solutions}} |
Revision as of 21:57, 21 April 2014
Contents
[hide]Problem
(Nazar Agakhanov, Russia) Let be positive real numbers such that . Prove that
Solution
Solution 1
We make the substitution , , . Then Since and are similarly sorted sequences, it follows from the Rearrangement Inequality that By the Power Mean Inequality, Symmetric application of this argument yields Finally, AM-GM gives us as desired.
Solution 2
We make the same substitution as in the first solution. We note that in general, It follows that and are similarly sorted sequences. Then by Chebyshev's Inequality, By AM-GM, , and by Nesbitt's Inequality, The desired conclusion follows.
Solution 3
Without clever substitutions: By Cauchy-Schwarz, Dividing by gives by AM-GM.
Solution 4
Proceed as in Solution 1, to arrive at the equivalent inequality But we know that by AM-GM. Furthermore, by Cauchy-Schwarz, and so dividing by gives as desired.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.